If it takes 25 mL of 0.05 M $H C l$ $HCl$ to neutralize 345 mL of $N a O H$ $NaOH$ solution, what is the concentration of the $N a O H$ $NaOH$ solution?

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If it takes 25 mL of 0.05 M $H C l$ $HCl$ to neutralize 345 mL of $N a O H$ $NaOH$ solution, what is the concentration of the $N a O H$ $NaOH$ solution?

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Answer 1 · 2016-06-27T05:41:45

$[N a O H] ≅ 4 \times 10^{- 3} \cdot m o l \cdot L^{- 1}$ $[NaOH]~=4xx10^-3*mol*L^-1$

Explanation:

$N a O H (a q) + H C l (a q) \to N a C l (a q) + H_{2} O (l)$ $NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)$

There is 1:1 equivalence between acid and base.

Thus $[N a O H]$ $[NaOH]$ $=$ $=$ $\frac{25 \times 10^{- 3} L \times 0.05 \cdot m o l \cdot L^{- 1}}{345 \times 10^{- 3} L}$ $(25xx10^-3Lxx0.05*mol*L^-1)/(345xx10^-3L)$

$=$ $=$ $? ? \cdot m o l \cdot L^{- 1}$ $??*mol*L^-1$

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If it takes 25 mL of 0.05 M $H C l$ $HCl$ to neutralize 345 mL of $N a O H$ $NaOH$ solution, what is the concentration of the $N a O H$ $NaOH$ solution?

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If it takes 25 mL of 0.05 M $H C l$ $HCl$ to neutralize 345 mL of $N a O H$ $NaOH$ solution, what is the concentration of the $N a O H$ $NaOH$ solution?